open generic points of affine scheme? Let $S=\text{Spec} A$ be an affine scheme, we assume $A$ is not a field, then we know irreducible components of $S$ correspond to all minimal prime ideals of $A$, in fact, these prime ideals are generic points of irreducible components of $S$.
Then I want to know if we assume all these generic points are open points in $S$(or at least one generic point is open), what property of $A$ can we deduce?
And what I mean by an open point is that this point is open in the subspace topology.
Thanks!
 A: For simplicitly let us assume that $\mathrm{Spec}(A)$ is integral (I'll leave you to think about the non-integral case) with unique generic point $\eta$. If $\{\eta\}$ is open then we know that there exists some neighborhood $D(f)$ of $\eta$ contained in $\{\eta\}$ and thus, of course, $D(f)=\{\eta\}$. From this, we see that
$$A[f^{-1}]=\mathcal{O}(D(f))=\mathcal{O}_{\mathrm{Spec}(A),\eta}=\mathrm{Frac}(A)$$
where the middle equality holds since there are no　neighborhoodsof $\eta$ properly contained in $D(f)$.
Conversely, we see that if there exists some $f$ in $A$ such that $\displaystyle A[f^{-1}]=\mathrm{Frac}(A)$ then we see that, in particular, $A[f^{-1}]$ is a field and so $\mathrm{Spec}(A[f^{-1}])$ only consists of one point. But, the map $\mathrm{Spec}(A[f^{-1}])\to \mathrm{Spec}(A)$ is an open embedding with image $D(f)$ and so, in particular, its image contains $\eta$. But, since $D(f)$ consists of only one point we must have that $D(f)=\{\eta\}$ and thus $\{\eta\}$ is open.
Thus, from the above we deduce the following:

Proposition: Let $A$ be a domain. Then the (unique) generic point $\eta$ of $\mathrm{Spec}(A)$ is open if and only if there exists some $f$ in $A$ such that $A[f^{-1}]=\mathrm{Frac}(A)$.

Let us give some simple examples/non-examples:
Example 1: Let $\mathcal{O}$ be a DVR with uniformizer $\pi$. Then, $\mathrm{Frac}(\mathcal{O})=\mathcal{O}[\pi^{-1}]$ and so you see that the generic point of $\mathrm{Spec}(\mathcal{O})$ is open. In fact, $\mathrm{Spec}(\mathcal{O})$ consists, as is used very often, of an open generic point $\eta$ and a closed point $(\pi)$.
Remark 1: More generally, if $K$ is a field and $\mathcal{O}$ is a so-called microbial valuation ring in $K$ (e.g. see [1, §I.1.5]) then $K=\mathcal{O}[\varpi^{-1}]$ for any pseudo-uniformizer $\varpi$ (e.g. see [1, Lemma I.1.5.9]) and so the generic point of $\mathrm{Spec}(\mathcal{O})$ is open. Such valuation rings play a pivotal role in Huber's theory of adic spaces. As an example, one can consider the the valuation induced on $\mathrm{Frac}(\mathbb{C}_p\langle t\rangle)$ by the valuation
$$\left|\sum_{n=0}^\infty a_n t^n\right|=\sup |a_n|$$
on
$$\mathbb{C}_p\langle t\rangle:=\left\{\sum_{n=0}^\infty a_n t^n:\lim a_n=0\right\}$$
Then, the valuation ring in $\mathrm{Frac}(\mathbb{C}_p\langle t\rangle)$ is an example of a microbial valuation ring. You can make even more exotic (non-rank $1$) examples. See [1, §I.1.5] again).
Non-example 2: Certainly $\mathbb{Z}$ doesn't have open generic point since there is no element $f$ in $\mathbb{Z}$ such that $\mathbb{Z}[f^{-1}]=\mathbb{Q}$. Indeed, this is clear by thinking about the fact that $v_p(f)\ne 0$ for only finitely many $p$ (where $v_p$ is the $p$-adic valuation).
Example/Non-example 3: If $A$ is finite type over a field $k$ (and a domain) then the generic point of $\mathrm{Spec}(A)$ is open if and only if $A$ is a finite extension of $k$. Indeed, one easy way to see this is that this implies that there exists some $f$ in $A$ such that $A[f^{-1}]=\mathrm{Frac}(A)$. Since $\mathrm{Spec}(A[f^{-1}])\to \mathrm{Spec}(A)$ is an open embedding this implies (e.g. see [2, Theorem 5.22(3)]) that $\mathrm{Spec}(A)$ has dimension zero from where the conclusion follows (e.g. see [2, Corollary 5.21]).
Remark 2: Combining Example 1 and Example/Non-example 3 we are able to observe an interesting subtelty. Namely, as we used in the latter (and is well-known) if $X$ is an irreducible variety then $\dim(U)=\dim(X)$ for any open subset $U$ of $X$. This is false for general rings as Example 1 shows since $\dim\{\eta\}=0$ but $\dim \mathcal{O}=1$ (in the DVR case).
References:
[1] Morel, S., 2019. Adic Spaces. Lecture Notes. https://web.math.princeton.edu/~smorel/adic_notes.pdf.
[2] Görtz, U. and Wedhorn, T., 2010. Algebraic geometry. Wiesbaden: Vieweg+ Teubner.
